I have to proof by contradiction that $ A \subseteq A $; if $ A \nsubseteq A $ then $ \exists x \in A ( x \notin A ) $ but $ \exists x \in A ( x \notin A ) $ is contradiction (in fact: $ \exists x ( x \in A \wedge x \notin A ) $) therefore $ A \subseteq A $ is true... The process is correct? Thank you all in advance

I don't think I agree. This just encourages you to believe that things are different when they are actually the same. The contradiction is clearer if you write "$x\in A$ and $x\notin A$" than if you write "$x\in A_1$, $x\notin A_2$ and $A_1=A_2$". But maybe there are other reasons to prefer this notation.
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Matthew PresslandMar 14 '13 at 15:42

@MattPressland: I know it might make some points you noted but if I wanted to prove like the OP did, I would prefered student find out what $A$ I mean when I write $x\in A$, the first $A$ or the second one. However, they both are the same.
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Babak S.Mar 14 '13 at 15:45

That is also reasonable. My feeling to a degree though is that the contradiction arises from the fact that when you write $x\in A$, you can't tell "which $A$" you mean - you have to mean both - so you can't possibly also have $x\notin A$.
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Matthew PresslandMar 14 '13 at 15:48

@MattPressland: Sometimes, I feel need a blackboard here to say what I mean.
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Babak S.Mar 14 '13 at 15:54

It's fine, this is just a semantic argument anyway. Different people will find different notation clearer.
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Matthew PresslandMar 14 '13 at 15:57